question_answer

The molar conductivity of \[0.25\,mol\,{{L}^{-1}}\] formic acid is \[40.1S\,c{{m}^{2}}mo{{l}^{-\,1}}.\]Calculate its dissociation constant. \[\begin{align}   & [\text{Assuming},\Lambda _{m({{H}^{+}})}^{{}^\circ }=349.6\,S\,c{{m}^{2}}mo{{l}^{-1}}; \\  & \Lambda _{(m)(HCO{{O}^{-}})}^{{}^\circ }=54.6\,S\,c{{m}^{2}}mo{{l}^{-1}}] \\ \end{align}\]

A) \[2.73\times {{10}^{-2}}\,mol\,{{L}^{-1}}\]    

B) \[3.74\times {{10}^{-\,3}}\,mol\,{{L}^{-1}}\]

C) \[4.75\times {{10}^{-\,3}}\,mol\,{{L}^{-1}}\]

D) \[5.27\times {{10}^{-\,3}}\,mol\,{{L}^{-1}}\]

Correct Answer: A

Solution :

\[\Lambda _{m}^{c}=40.1\,\,\,S\,\,\,c{{m}^{2}}\,\,mo{{l}^{-1}}\]

\[\Lambda _{m}^{{}^\circ }=\Lambda _{m(HCO{{O}^{-}})}^{{}^\circ }+\Lambda _{m({{H}^{+}})}^{{}^\circ }\]

\[=(54.6+349.6)S\,c{{m}^{2}}mo{{l}^{-\,1}}\]\[=404.2\,S\,c{{m}^{2}}mo{{l}^{-\,1}}\]

\[\alpha =\frac{\Lambda _{m}^{c}}{\Lambda _{m}^{{}^\circ }}=\frac{40.1}{404.2}=0.0992\]

Now,

\[{{K}_{\alpha }}=\frac{C{{\alpha }^{2}}}{1-\alpha }\] \[=\frac{0.25\times {{(0.0992)}^{2}}}{(1-(0.0992)}\]\[=2.73\times {{10}^{-\,3}}mol\,{{L}^{-\,1}}\]